I have the following statement to prove:
Let $V$ be a normed vector space (so not necessarily complete or finite dimensional) over $\mathbb{R}$. Take any norm $\Vert \cdot \Vert_V$ on $V$ and let $\Vert \cdot \Vert_{V\rightarrow V}$ be the induced norm. Suppose $f : V \rightarrow V$ is bounded and $\Vert f \Vert_{V\rightarrow V} \leq \beta < 1$. If $(id - f)^{-1} $ exists, show that $\Vert (id - f)^{-1} \Vert_{V\rightarrow V} \leq \frac{1}{1-\beta}$, where $id :V \rightarrow V$ is the identity operator.
My attempt:
First by reverse triangle inequality, we get $$ \Vert id - f \Vert_{V \rightarrow V} \geq \Vert id \Vert_{V \rightarrow V} - \Vert f\Vert_{V \rightarrow V} \geq 1 - \beta \implies \frac{1}{\Vert id - f \Vert_{V \rightarrow V}} \leq \frac{1}{1 - \beta}. $$
Then by submultiplicity of the induced norm, we get $$ 1 = \Vert(id - f)(id - f)^{-1} \Vert_{V \rightarrow V} \leq \Vert id - f \Vert_{V \rightarrow V}\Vert (id - f)^{-1} \Vert_{V \rightarrow V}, $$ which implies $$ \Vert (id - f)^{-1} \Vert_{V \rightarrow V} \geq \frac{1}{\Vert id - f \Vert_{V \rightarrow V}} \leq \frac{1}{1-\beta}. $$
So this attempt ended up with an inconclusive result. I'm guessing I made a silly mistake somewhere in the inequalities... Any hint on what I did wrong here?
More simply, if $y = (I-f)^{-1}(x)$, then \begin{equation} (I - f)(y) = x \Rightarrow y = x + f(y)\Rightarrow \|y\|\le\|x\| + \beta \|y\| \end{equation} Hence \begin{equation} (1-\beta)\|y\|\le\|x\|\quad\Rightarrow\quad \|(I-f)^{-1}(x)\| = \|y\|\le\frac{1}{1-\beta}\|x\| \end{equation}